Question 251713
We know the bank has nickels and dimes and that there are 46 coins
n+d=46
The number of nickels and dimes adds up to 46

5(n-2)+10(d-5)=340
Now that we are talking dollar amounts we are taking the value of the coins.
So if I take away 2 nickels i am taking away 10 cents and if I take away 5 dimes I am taking away 50 cents
and after all that is done I have $3.40 cents
Each nickel is worth 5 cents and of course each dime is 10 cents
5n is the value of the nickels and 10d is the value of the dimes.
but we don't know how much there was but how much there is now after some were removed.
so we can say there are 7 fewer coins =39
and 5n+10d=340
n+d=39 
So there would be 29 dimes and 10 nickels NOW

or we can say
n+d=46
5(n-2)+10(d-5)=340

See how we are removing two nickels and 5 dimes
-5(2) and 5 dimes -5(10)
and we find that there WERE 34 dimes and 12 nickels
So I did the problem two different ways and came up with the results for before and after
The problem doesn't really say if it wants to know NOW or BEFORE.